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1.4.7
.
Finish the following proof for Theorem
1.4.12.
Assume
B
is a countable set. Thus, there exists
f
:
N
!
B
,
which is
and onto. Let
A
B
B
.
We must show that
A
is countable.
Let
n
1
= min
f
n
2
N
:
f
(
n
)
2
A
g
.
g
:
N
!
A
,
set
g
(1) =
f
(
n
1
)
.
Show how to inductively continue this process to
produce a
function
g
from
N
onto
A
.
Shouldn±t one say, ²Show how to continue this process inductively³?
Notice that
f
n
2
N
:
f
(
n
)
2
A
g
is
f
1
(
A
)
, so we can write
n
1
= min
f
1
(
A
)
.
Then write
n
2
= min
n
2
f
1
(
A
) :
n > n
1
±
n
3
= min
n
2
f
1
(
A
) :
n > n
2
±
and, in general,
n
i
+1
= min
n
2
f
1
(
A
) :
n > n
i
±
That±s the construction. Because
A
is in´nite, and
f
maps onto
B
, the set
of natural numbers
n
2
f
1
(
A
) :
n > n
i
±
is nonempty no matter what the natural number
n
i
is. So this set has a
minimum element. Clearly
n
1
< n
2
< n
3
<
±±±
by the way we chose them.
We de´ne
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 Spring '10
 MetCalfe

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